Re: "Embedded" in the plane.

From: W. Dale Hall (mailtowd_hall_at_pacbell.net)
Date: 07/27/04 

Date: Tue, 27 Jul 2004 02:13:12 GMT

Bill Jones wrote:

> harrisq@tcs.inf.tu-dresden.de (Mitch Harris) wrote in message news:<2mict3Fnb0csU3@uni-berlin.de>...
>
>>Bill Jones <bill_jones92057@yahoo.com> wrote:
>>
>>>Doesn't 'embedded' in the plane connote some thickness to the plane?
>>
>>Why would it?
>>
>>Mitch
>
>
> Do you know what 'connote' means?

Sorry to *** in here, but I do know what connote means, yet do not have
any sense of why "embedded in the plane" should connote some thickness
to the plane.

In the standard (to me), topological, sense, a space X is embedded in
another space Y if X is mapped into Y as a homeomorphism onto its image.
For instance, the circle S^1 is embedded in the plane by the mapping
theta |--> (cos(theta), sin(theta)).

As a contrary example (to something not being embedded, the real
line is not embedded in the plane by the mapping that setds 0 to the
center point of a figure 8, wraps the positive reals around the top
loop (asymptotically approaching the center point as x --> infinity)
counterclockwise, and wraps the negative reals similarly around the
bottom loop. While it is 1:1 and surjective (to the figure 8), the
mapping fails to be invertible at the center point, as the two ends
of the line are mapped to neighborhoods of that point.

The standard inclusions of the real line into the plane (e.g., the
x and y axes in the Cartesian coordinate system) are indeed embeddings.

There is a bit of ambiguity available to the term "thickness", which
I'll attempt to dispel by interpreting it as referring to a third
dimension. In that case, the plane has no thickness.

Given all that, why should "embedded" in the plane connote some
thickness to the plane?

Dale.

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